At 298 K temp. {K_{sp}} of Ca{F₂} is 1.7 × {10^{ - 10}} . One person daily drinks 2.5 L saturated

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6-2.Equilibrium-II (Ionic Equilibrium)

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At $298$ $K$ temp. the ${K_{sp}}$ of $Ca{F_2}$ is $1.7 \times {10^{ - 10}}$. One person daily drinks $2.5$ $L$ saturated water by $Ca{F_2}$ . Then how much gram $Ca{F_2}$ present in his body? (Molecular mass of $Ca{F_2}$ is $ 78$ $g$ $mol^{-1}$ )

Option A

Option B

Option C

Option D

Solution

$\mathrm{S}=3.489 \times 10^{-4} \mathrm{M}$ and $0.0683 \mathrm{~g}$

Standard 11

Chemistry

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